Problem 150 Pointsprint Media Advertising Pma Has Been Given

Problem 150 Pointsprint Media Advertising Pma Has Been Given A Con

Print Media Advertising (PMA) has been assigned a contract to market Buzz Cola through newspaper advertisements in a major Southern newspaper. The costs are $2000 for a full-page ad on each weekday (Monday through Saturday), and $8000 for a full-page ad on Sundays. The daily circulation is 30,000 copies on weekdays and 80,000 on Sundays. PMA has a total advertising budget of $40,000 for August. They wish to run at least eight weekday ads and at least two Sunday ads during the month. Partial ads are permissible, meaning a fractional ad (e.g., 3.5 ads) corresponds to three full ads and one half ad, with costs and exposure reducing proportionally. August has 26 weekdays and 5 Sundays. The goal is to maximize total exposure, measured by circulation, over August.

Paper For Above instruction

a) Formulate the linear programming model for this problem.

The decision variables are the number of weekday ads and Sunday ads, possibly fractional, to be placed during August:

  • Let x = total number of weekday ads (including fractional ads) to be placed during August.
  • Let y = total number of Sunday ads (including fractional ads) during August.

The objective function aims to maximize the total exposure:

Maximize Z = 30,000 x + 80,000 y

Subject to the budget constraint:

2000 x + 8000 y ≤ 40,000

Minimum ad constraints:

x ≥ 8 (at least eight weekday ads)

y ≥ 2 (at least two Sunday ads)

Ad placement limits based on days available:

0 ≤ x ≤ 26

0 ≤ y ≤ 5

Additionally, fractional ads are allowed, reflecting partial placement, and costs and exposure are proportionate to the fractional ads.

b) Find the optimal solution using the graphical method. Show all steps, including algebraic points of intersection.

To solve graphically, plot the constraints:

  • The budget constraint: 2000x + 8000y = 40,000
  • The minimum ad constraints: x ≥ 8, y ≥ 2
  • Maximum ads based on days: x ≤ 26, y ≤ 5

Convert the budget constraint to slope-intercept form:

2000x + 8000y = 40,000

⇒ 8000y = 40,000 - 2000x

y = (40,000 - 2000x) / 8000 = 5 - 0.25x

Next, identify corner points:

  1. At x = 8 (minimum weekday ads): y = 5 - 0.25*8 = 5 - 2 = 3.
  2. At y = 2 (minimum Sunday ads): x = (40,000 - 8000*2) / 2000 = (40,000 - 16,000) / 2000 = 24,000 / 2000 = 12.
  3. At maximum daily limits: x = 26 (max weekdays), then y = 5 (max Sundays): check budget:

    200026 + 80005 = 52,000 + 40,000 = 92,000 > 40,000 → not feasible. So, the allocated ads at maximum days are not feasible due to budget constraint.

Find feasible intersection point between x and y along the budget constraint within the permissible bounds:

At point x = 8: y = 3; total exposure = 30,0008 + 80,0003 = 240,000 + 240,000 = 480,000

At point x = 12: y = 2; total exposure = 30,00012 + 80,0002 = 360,000 + 160,000 = 520,000

At the intersection point where 2000x + 8000y = 40,000, with y ≥ 2 and x ≥ 8:

Choosing the point with the maximum exposure: at (x,y) = (12, 2), the exposure is 520,000, which is optimal within the constraints.

Thus, the optimal fractional solution would be approximately at (x,y) = (12, 2). Adjusted if fractional placement is allowed, the exposure would be maximized at this point.

c) Find the optimal solution using Excel Solver. Copy and paste the Excel spreadsheet and the Answer Report.

In Excel, set up cells to represent variables x and y, along with formulas for total cost and exposure. Use Solver to maximize exposure subject to the constraints specified. The solution will confirm the algebraic findings: the optimal point at approximately x = 12, y = 2, with total exposure of about 520,000. The detailed spreadsheet and solver report can be generated in Excel, following the model outlined above.

References

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